Problem 3 (Basis for the kernel of a matrix) Let A be an mxn matrix and consider W = {x = Rn | Añ = 0}. We know from assignment 1 that W is a subspace of Rn. In this problem, we wish to explain why the fol- lowing strategy to find a basis of W works: if the solutions to A = 0 are given by = t₁v₁ + ··· + tk√k, where t1,..., tk are free parameters, then the vectors 1,..., Uk form a basis of W. (a) Show that if the matrix A' is obtained from the matrix A by a row operation (i.e. adding a multiple of a row to another row, multiplying a row by a nonzero constant or swapping two rows), then {ã € Rª | Añ = 0} = {ã € R″ | A'ñ = 0}. Deduce that if B is the row-reduced echelon form of A, then {ã € R” | Añ = 0} = {ã € R″ | Bì = 0}. (b) Looking at the row-reduced echelon form of A, we can see which variables are "leading", i.e. have a leading 1 in their corresponding column, and which variables are "free", i.e. have no leading 1 in their corresponding column. When solving the system B = 0, we see that each leading variable can be expressed in terms of the free variables that succeed it. In other words, if x, is a leading variable, then we have n xi = Σ Gijli 1 j=i+1 xj free for some constants Cij E R. Let i₁ < ... < ik denote the indices of the free variables. Since the free variables xi, really are "free" to have any value t;, we conclude that the solutions to the system BÃ = Ō are given by x1 Ti1-1 Ti1 Ti+1 1 = = t1 ++tk Tik-1 Tik Tik+1 In Uk √1 Prove that the vectors 1,..., Vk are linearly independent. (c) Conclude that {1,..., k} is a basis of W. Remark: To help digest all the notational intricacies involved in this problem, it might be helpful to work in parallel with the particular matrix 1 2 3 4 A=0 0 1 2 00 2 4
Problem 3 (Basis for the kernel of a matrix) Let A be an mxn matrix and consider W = {x = Rn | Añ = 0}. We know from assignment 1 that W is a subspace of Rn. In this problem, we wish to explain why the fol- lowing strategy to find a basis of W works: if the solutions to A = 0 are given by = t₁v₁ + ··· + tk√k, where t1,..., tk are free parameters, then the vectors 1,..., Uk form a basis of W. (a) Show that if the matrix A' is obtained from the matrix A by a row operation (i.e. adding a multiple of a row to another row, multiplying a row by a nonzero constant or swapping two rows), then {ã € Rª | Añ = 0} = {ã € R″ | A'ñ = 0}. Deduce that if B is the row-reduced echelon form of A, then {ã € R” | Añ = 0} = {ã € R″ | Bì = 0}. (b) Looking at the row-reduced echelon form of A, we can see which variables are "leading", i.e. have a leading 1 in their corresponding column, and which variables are "free", i.e. have no leading 1 in their corresponding column. When solving the system B = 0, we see that each leading variable can be expressed in terms of the free variables that succeed it. In other words, if x, is a leading variable, then we have n xi = Σ Gijli 1 j=i+1 xj free for some constants Cij E R. Let i₁ < ... < ik denote the indices of the free variables. Since the free variables xi, really are "free" to have any value t;, we conclude that the solutions to the system BÃ = Ō are given by x1 Ti1-1 Ti1 Ti+1 1 = = t1 ++tk Tik-1 Tik Tik+1 In Uk √1 Prove that the vectors 1,..., Vk are linearly independent. (c) Conclude that {1,..., k} is a basis of W. Remark: To help digest all the notational intricacies involved in this problem, it might be helpful to work in parallel with the particular matrix 1 2 3 4 A=0 0 1 2 00 2 4
Linear Algebra: A Modern Introduction
4th Edition
ISBN:9781285463247
Author:David Poole
Publisher:David Poole
Chapter2: Systems Of Linear Equations
Section2.3: Spanning Sets And Linear Independence
Problem 31EQ
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Transcribed Image Text:Problem 3 (Basis for the kernel of a matrix) Let A be an mxn matrix and consider W = {x = Rn | Añ =
0}. We know from assignment 1 that W is a subspace of Rn. In this problem, we wish to explain why the fol-
lowing strategy to find a basis of W works: if the solutions to A = 0 are given by = t₁v₁ + ··· + tk√k, where
t1,..., tk are free parameters, then the vectors 1,..., Uk form a basis of W.
(a) Show that if the matrix A' is obtained from the matrix A by a row operation (i.e. adding a multiple of a
row to another row, multiplying a row by a nonzero constant or swapping two rows), then
{ã € Rª | Añ = 0} = {ã € R″ | A'ñ = 0}.
Deduce that if B is the row-reduced echelon form of A, then
{ã € R” | Añ = 0} = {ã € R″ | Bì = 0}.
(b) Looking at the row-reduced echelon form of A, we can see which variables are "leading", i.e. have a
leading 1 in their corresponding column, and which variables are "free", i.e. have no leading 1 in their
corresponding column. When solving the system B = 0, we see that each leading variable can be expressed
in terms of the free variables that succeed it. In other words, if x, is a leading variable, then we have
n
xi =
Σ Gijli
1
j=i+1
xj free
for some constants Cij E R. Let i₁ < ... < ik denote the indices of the free variables. Since the free
variables xi, really are "free" to have any value t;, we conclude that the solutions to the system BÃ = Ō
are given by
x1
Ti1-1
Ti1
Ti+1
1 =
= t1
++tk
Tik-1
Tik
Tik+1
In
Uk
√1
Prove that the vectors 1,..., Vk are linearly independent.
(c) Conclude that {1,..., k} is a basis of W.
Remark: To help digest all the notational intricacies involved in this problem, it might be helpful to work in
parallel with the particular matrix
1 2 3 4
A=0 0 1 2
00 2 4
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